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32. (5.1, 5.4) Newton's second law 2 F41 1 4 F43 3 F42 a F net In an inertial reference frame, the acceleration of a particle is proportional to the net force (the sum of all forces) exerted on the particle and inversely proportional to the mass of the particle: r r r Fnet = ∑ Fi = ma i =all (the SI unit of force is 1N=1kg⋅1m/1s2) 33. (5.6) Newton's third law If one body exerts a force on another body, the second body exerts an opposite force on the first one: r r F12 = − F21 F 12 F 21 1 2 41 34. (6.6, 5.1) Mechanisms of interaction (forces) On the microscopic scale there are only four kinds of interaction between particles: gravitational (attractive), electromagnetic (attractive or repulsive), strong nuclear force, weak nuclear force. To simplify the analysis of interactions between bodies (systems of particles) it is convenient to introduce complex interactions. 35. (14.1) The gravitational force a) A particle with mass m1, F 21 separated from a particle with 1 r12 mass m2 exerts r an attractive 2 force (vector) F21 on the other particle r F21 = − G m12m2 ⋅ r$12 r12 The symbol r G denotes the universal gravitational constant, and r$12 = r12 a unit vector in the relative position of the r12 particles. b) Objects with spherically distributed mass, also obey the above equations with r representing the distance between the centers of the objects. The force is exerted toward the center of the object exerting the force. 42 36. (5.5, 14.3) Weight The product of the mass of an object and the free fall acceleration at the location of the object is called the weight of the object: r r W = mg If an objects moves near the surface of a planet, then its distance to the center of the planet, for all practical purposes, does not change. On a planet with radius R and mass M the weight of a body is approximately equal to the gravitational force exerted on the object by the planet: W = GM m R2 Therefore g = GM R2 On earth g = 9.80 m/s2. 37. (5.6, 12.4) The normal force W r The normal force F N is the N vector component of a force that a rigid surface exerts (due to deformation; strain) on an object with which it is in contact, namely, in the direction perpendicular to the surface. The normal force prevents objects from crossing the surface, therefore it is dependent on the other forces applied to the body. 43 38. (5.8) Static frictional force F Static frictional force is a component of a force that a rigid surface exerts (due to strain) on fs the surface of an object with which it is in contact, namely, in the direction parallel to the surface. Within certain limits the static friction (interaction) prevents the object's surface from moving along the rigid surface f s ≤ µs N N The coefficient µs is called the coefficient of static friction. Static frictional force depends on the other forces applied to the body. 39. (5.8) Kinetic frictional force Kinetic frictional force is a component of a force that a rigid surface exerts (due to deformation) on the surface of an object with which it is in contact, namely, in the direction parallel to the surface. fk = µk N The coefficient µk is called the coefficient of kinetic friction. The kinetic frictional force is always in the direction opposite to the relative velocity of the surfaces. 44 40. (5.7, 12.4) Tension force T -T r Tension force T is a force that a surface exerts (due to deformation; strain) on an object with which it is in contact, namely, in the direction perpendicular to the surface. Tension force prevents the object from leaving the surface, therefore it is dependent on the other forces applied to the body. 41. (5.7) Free-Body Diagram In order to successfully predict the motion of a body, one has to recognize all the forces acting on the body. A figure with all the forces marked is called a free body diagram. In identifying the forces affecting the motion of the body make sure that the forces are exerted on the considered body. Also remember that each force has its source in another body. When using the scalar components of vector quantities related to a certain body be consistent with the choice of the reference frame. (If two or more bodies are discussed in a single problem, they can be considered in separate reference frames.) 45 42. (6.3) Motion in noninertial reference frames y' Let's relate the scalar components of velocity and Ω y acceleration in case of an arbitrary motion of the primed r' x' reference frame with respect to O' the unprimed reference frame. r R z' The primed reference frame can x move, rotate and accelerate, but O z let's maintain that time passes by identically in both reference frames. In the unprimed reference frame we change constantly the base ˆi ', ˆj', kˆ ' of the unprimed reference frame. Notice v that all three base vectors rotate with angular velocity Ω , therefore the time derivatives of those base vectors represented in the unprimed reference frame are dˆi' r ˆ dˆj' r ˆ dkˆ ' r ˆ i) = Ω × i', = Ω × j' , = Ω × k' dt dt dt If in the primed reference frame the components of the position vector are r ' (t ) = [x ' (t ), y ' (t ), z' (t )] then the position vector in the unprimed reference frame is r ii) r '( t ) = x ' (t )ˆi ' (t ) + y' (t )ˆj' (t ) + z' (t )kˆ ' (t ) j' k' i' j i k { } Therefore, we can relate the position vector (oriented segment) in the unprimed reference frame in terms of the scalar components in the primed reference frame and the primed base: r r iii) r( t ) = R (t ) + x ' (t )ˆi ' (t ) + y' (t )ˆj' (t ) + z' (t )kˆ ' (t ) 46 In order to relate velocity in both frames we have to find calculate the time derivative of both sides: r r r dr( t ) dR dx ' ˆ dy' ˆ dz' ˆ dˆi ' dˆj' dkˆ ' = v(t ) = = + i'+ j'+ k ' + x' + y' + z' dt dt dt dt dt dt dt dt r r r r = V ( t ) + v' x ˆi '+ v' y ˆj'+ v' z kˆ ' + x ' Ω × ˆi ' + y' Ω × ˆj' + z' (Ω × kˆ ') = r r r v = V ( t ) + v' x ˆi '+ v' y ˆj'+ v' z kˆ ' + Ω × x ' ˆi ' + Ω × y' ˆj' + (Ω × z' kˆ ') = r r = V ( t ) + v' x iˆ'+ v' y ˆj'+ v' z kˆ ' + Ω × x' ˆi'+ y' ˆj'+ z' kˆ ' ( ( ( iv) ) ( ) ( ) ( ) ( ) ( ( ) ) )) r r r r r v (t ) = V( t ) + v ' (t ) + (Ω(t ) × r ' (t )) In order to find the relation between acceleration we have to calculate the time derivative of velocity. ( ( ) ( )) r r d r a(t ) = V + v' x ˆi '+ v' y ˆj'+ v'z kˆ ' + Ω × x' ˆi '+ y' ˆj'+ z' kˆ ' = dt r dV dv' x ˆ dv ' y ˆ dv' z ˆ dx ' r ˆ dy' r ˆ dz ' r ˆ (Ω × k') + = + i'+ j'+ k ' + Ω × i' + Ω × j' + dt dt dt dt dt dt dt r r dx ' dy' dz' r r r dΩ ˆj'+ kˆ ' + x' Ω × iˆ' + y' Ω × ˆj' + z' (Ω × kˆ ') = + × x' ˆi'+ y' ˆj'+ z' kˆ ' + Ω × iˆ'+ dt dt dt dt r r dV = + a' x ˆi '+ a' y ˆj'+ a' z kˆ ' + 2 ⋅ Ω × v' x ˆi '+ v' y ˆj'+ v' z kˆ ' + dt r r r dΩ + × x' ˆi'+ y' ˆj'+ z' kˆ ' + Ω × Ω × x' ˆi'+ y' ˆj'+ z' kˆ ' dt ( ( v) ( ) ( ( ) ( ( ) ) ) ( ) ( ) ( ( )) r r r r r r r r r r a = A + a'+ 2Ω × v '+ α × r '+ Ω × (Ω × r ') 47 ) ) Let's assume that the unprimed reference frame is an inertial reference frame. The net force of all the force exerted on a particle by other bodies is equal to the product of the particle's mass and acceleration (in unprimed reference frame): r r vi) Fnet = ma However, in the primed reference frame product of mass and acceleration is equal to the sum of the net force exerted on the particle and a few other terms! r r r r r r r r r r vii) ma' = Fnet − mA − 2 mΩ × v '− mα × r − mΩ × (Ω × r ') Each of the terms represent a certain kind of a fictitious "force" (also called "inertial force"). Notice that other bodies do not exert these “forces”. 48 WORK AND ENERGY 43. (7.1, 7.3) Work (scalar) performed on an object by another object (by the body exerting the force) is a measure of cumulative interaction. It depends on the force and the path of a particular point of the body. Consequently it depends on the choice of the reference frame. a) The work dW done on a particle displaced along r differential pathr dr, by an object exerting force F is defined as F A dr B r r dW ≡ F ⋅ dr r r W F We can define work in an integral form = ∫ ⋅ dr . path The SI unit of work is 1J=1N⋅1m b) For the work done on an object, one must specify on which point associated with the object the work is done (the center of mass, the point of the force application ... ). 49 Example 1: Work done by the gravitational force (near the earth's surface) on the center of mass of an object with mass m is r r Wg = ∫ W ⋅ dr = ∫ − mg ⋅ dy = − mg ( h f − h i ) = − mg ∆h path path where ∆h is the change in the object's altitude. Example 2: Force independent of position r r r r r r W = ∫ F ⋅ dr = F ⋅ ∫ dr = F ⋅ ∆r path path Example 3. Work done by a spring being stretched. Fx = -kx x 0 x The force created in a spring is related to the deformation of the spring by Hooke's law: r r r F(x ) = − kx r where the position vector x represents the displacement of the free end from the relaxed position in the direction of the spring. The proportionality coefficient k is called the spring constant. Work done by the spring is therefore r r x kx 2 W = ∫ F ⋅ d r = ∫ (− kx )dx = − 2 path 0 50 44. (7.4) Definition of the kinetic energy (classical) A particle with mass m, moving with speed v has kinetic energy K of mv 2 K≡ 2 (The SI unit of kinetic energy is 1J) Note. Kinetic energy is used to describe the motion of an object. When an object is approximated by a particle, the kinetic energy defined above is called the translational kinetic energy of the object. 45. (7.5) The Work-Energy Theorem I (for the net force of all forces exerted on a particle). In an inertial reference frame, the work dW done by all the forces exerted on the particle (the net force) is equal to the change in the kinetic energy dK of the particle dW = dK (or ∆ Wnet = ∆ K) The work-energy theorem is a direct consequence of the definition of kinetic energy, the definition of work and Newton's second law. proof. r r m( rv )2 r r r dv r r r = dK dWnet = Fnet ⋅ dr = ma ⋅ dr = m ⋅ vdt = mv ⋅ dv = d dt 2 51 Example: What is the speed of an object after falling 5m from a resting position? a) from Newton's 2nd law Assume the initial location of the object is the reference point and the y-axis is in the vertical direction; t0 = 0s t1 = ? - the time of the release; - the time corresponding to the instant when the object is at a position 5m below the release point; v0=0 m/s - the initial velocity of the object v1=? m/s - the velocity of the object at instant t1 a=[0,-g] - the acceleration (constant) of the object Only the y-direction provides nontrivial equations. v y ( t1 ) = 0 − 10 t1 2 − 10 t 1 5 = ( y( t1 )) = 0 + 0 ⋅ t1 + 2 From the second equation, we find that the object is 5m below the release point 1s after it was released. Using this value in the first equation we find that the vertical component of velocity is v y ( t1 ) = − 10 m / s2 Since the other components are zero, the speed of the object is 10m/s. 52 b) from the work-energy theorem K0 = 0J K1= ? m =? - the initial (translational) kinetic energy of the object - the final (translational) kinetic energy of the object - mass of the object Since the object is in the free fall motion, the only work performed on the object is the gravitational work (done by earth). The gravitational force is constant throughout the motion therefore the net work is Wnet = Wg = − mg ∆ h = − m ⋅ 10 m / s2 ⋅ ( − 5 m ) That work causes a change in the kinetic energy and according to the work-energy theorem I mv 22 mv12 − = − mg ∆h 2 2 Therefore v = − 2g∆ h = − 2 ⋅ 10 m / s2 ⋅ ( − 5 m ) = 10 m / s2 53 46. (7.5) Power a) Definition The power of a force is defined as the rate at which work is done by that force. dW P( t ) ≡ dt (The SI unit is 1W=1J/1s) b) From the definition it follows that the work ∆W done in a certain time interval ∆ t = t 2 − t1 is t2 ∆W = ∫ P( t )dt t1 c) The average power during this interval is t2 Pav = ∫ P(t )dt t1 t 2 − t1 = ∆W ∆t r d) When a force F is exerted by a body on a particle r moving with velocity v, the power delivered by this body is r r r r P = dW = F ⋅ dr = F ⋅ v dt dt 54 47. (8.intro., 8.2) Conservative and nonconservative forces B A a) (Definition) If the work done by a force on an object moving between two positions is independent of the path of the motion, the force is called a conservative force. All other forces are called nonconservative forces. b) (Theorem) The work done by a conservative force around a loop (the object returns to its initial position) is zero. 48. (8.2) Definition of potential energy If a force exerted on a particle is conservative, the change in potential energy dU from one position to another is defined by the work dW performed by that force as follows: dU ≡ - dW (or ∆U = -∆U) This definition assigns potential energy only with accuracy to a constant. We have to choose an arbitrary position of the object (the reference point for potential energy) to which we assign zero potential energy. 55 49. (8.1) Gravitational potential energy (near the surface) Ug The gravitational potential energy Ug is the potential energy that an object has, by virtue of position, above the surface of the earth. h The gravitational potential energy Ug of a particle with mass m, placed at a position with a vertical component different by h from the reference location is Ug = mgh proof. r yo +h r r r r r r Ug (r ) = Ug , ref − Wro →r = − r∫ mgdr = − ∫ − mgdy = mgh ro y0 50. (14.7) Gravitational potential energy (in general) The gravitational potential Ug energy of a particle with mass m, placed at distance r from another particle with mass M is r Mm UG (r ) = −G r (The reference point for the potential energy is far away from the object) proof. r Mm r Mm Mm 1 1 U G = − ∫ G 2 rˆ ⋅ dr = − ∫ − G 2 dr = −GMm − = −G r r r r ∞ path ∞ 56 r 51. (8.2) The ideal spring Fx = -kx x x 0 r a) The restoring force F of an ideal spring depends on the r displacement x, by which the spring is stretched or compressed from the unstrained position r r F = −kx (the coefficient k is called the spring constant.) Us b) The elastic potential energy U s that an ideal spring has by virtue of being stretched or compressed is 2 x U s = kx 2 where x is the change in length of the spring from its unstrained length. proof. kx 2 Us = − Ws = − ∫ − kx ' dx ' = 2 0 x 57 52. (8.4) Definition of mechanical energy; The sum of the kinetic and the potential energy of a particle is called the total mechanical energy of the particle. E≡K+U 53. (8.5) Work-energy theorem II If some forces exerted on a particle are conservative, the work Wnc, done by all forces not included in the potential energy, is equal to the change in the mechanical energy ∆E of the particle. Wnc = ∆ E proof. ∆ E = ∆K + ∆ U = Wnet − Wc = Wnc 58 Example: Is it safe for an m = 100 kg person to bungee-jump from a height of h = 200 m, if the bungee-cord has a length of l = 50 m and its spring constant is k = 20 N/m ? solution 1: work-energy theorem I h From the definition of potential energy we can express the work done by the gravitational force and the elastic force in terms of the change in position of the person U Wg = − ∆U g = mg (l + ∆l ) k ( ∆l ) 2 We = − 2 Ue Ug U (The second expression is valid only when the cord is stretched) If we ignore the resistance of air, no other forces are performing work on the person. The change in the kinetic energy of the person between the top of the tower and the lowest point of the jump is zero. Therefore from the work energy theorem we obtain a quadratic equation with one unknown ∆l. k ( ∆l ) 2 mg (l + ∆l ) − =0 2 k (∆l ) 2 − 2 mg∆l − 2 mg l = 0 ( This equation has two mathematical solutions 59 ) mg ± ( mg) 2 + 2kmgl ∆l = = k 100kg ⋅10m / s 2 ± (100kg ⋅10m / s 2 ) 2 + 2 ⋅ 20 N / m ⋅100kg ⋅10m / s 2 ⋅ 50m = = 20N / m 136 m = − 37m Only the positive solution is physically acceptable, because the formula for elastic potential energy Ue is valid only for stretching. At the lowest position the person will be at height y = 200m-(50m+136m) = 14 m above the ground. solution 2: work-energy theorem II From the definition of potential energy, we can express the work done by the elastic force in terms of the change in position of the person k ( ∆l ) 2 We = − 2 We can include the gravitational potential energy Ug into the mechanical energy E. During motion, the mechanical energy changes from its initial value Ei = 0 J + 0 J to its value at the lowest point Ef = 0J + mg [ − ( l + ∆ l )] Using the work-energy theorem II, we obtain the same quadratic equation k ( ∆ l )2 − = − mg ( l + ∆ l ) 2 60